Question: You have found the following ages (in years) of 6 zebras. The zebras are randomly selected from the 36 zebras at your local zoo: $ 11,\enspace 5,\enspace 1,\enspace 15,\enspace 17,\enspace 16$ Based on your sample, what is the average age of the zebras? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 36 zebras, we are only able to estimate the population mean and variance by finding the sample mean $({\overline{x}})$ and sample variance $({s^2})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6$ To compensate for this underestimation, rather than simply averaging the squared deviations from the mean , we total them and divide by $n - 1$ $ {s^2} = \dfrac{\sum\limits_{i=1}^{{n}} (x_i - {\overline{x}})^2}{{n - 1}} $ $ {s^2} = \dfrac{{0.04} + {33.64} + {96.04} + {17.64} + {38.44} + {27.04}} {{6 - 1}} $ $ {s^2} = \dfrac{{212.84}}{{5}} = {42.57\text{ years}^2} $ We can estimate that the average zebra at the zoo is 10.8 years old. There is a variance of 42.57 years $^2$.